Problem 97 - Chemistry: The Central Science

Question

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) \(\mathrm{Ni}^{+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{Ni}(s)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{2-}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s)\) (acidic solution)

Step-by-Step Solution

Answer

The balanced disproportionation reactions are: (a) \(\mathrm{2Ni}^+ \longrightarrow \mathrm{Ni}^{2+} + \mathrm{Ni}\) (b) \(\mathrm{2MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{4}^{-} + \mathrm{MnO}_{2} + 2H_2O + 4H^+\)

Step 1: Identify the half-reactions

First, we can identify the oxidation and reduction half-reactions. The oxidation half-reaction involves \(\mathrm{Ni}^+\) losing an electron to become \(\mathrm{Ni}^{2+}\), while the reduction half-reaction involves \(\mathrm{Ni}^+\) gaining an electron to become \(\mathrm{Ni}\) (solid). The half-reactions are: 1. Oxidation: \(\mathrm{Ni}^+ \longrightarrow \mathrm{Ni}^{2+} + e^-\) 2. Reduction: \(\mathrm{Ni}^+ + e^- \longrightarrow \mathrm{Ni}\)

Step 2: Balance the half-reactions

The oxidation half-reaction is already balanced. For the reduction half-reaction, we don't need to balance anything since the reaction involves only one species and one electron.

Step 3: Combine the half-reactions

Since both half-reactions involve transfer of one electron each, we can directly add them to obtain the balanced disproportionation reaction: \(\mathrm{2Ni}^+ \longrightarrow \mathrm{Ni}^{2+} + \mathrm{Ni}\) (b) Balancing the disproportionation reaction of \(\mathrm{MnO}_{4}^{2-}(aq) \longrightarrow \mathrm{MnO}_{4}^{-}(aq)+\mathrm{MnO}_{2}(s)\) (acidic solution):

Step 4: Identify the half-reactions

First, we can identify the oxidation and reduction half-reactions. The oxidation half-reaction involves \(\mathrm{MnO}_{4}^{2-}\) losing an electron to become \(\mathrm{MnO}_{4}^-\), while the reduction half-reaction involves \(\mathrm{MnO}_{4}^{2-}\) gaining an electron to become \(\mathrm{MnO}_{2}\) (solid). The half-reactions are: 1. Oxidation: \(\mathrm{MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{4}^- + e^-\) 2. Reduction: \(\mathrm{MnO}_{4}^{2-} + e^- \longrightarrow \mathrm{MnO}_{2}\)

Step 5: Balance the half-reactions

The oxidation half-reaction is already balanced. For the reduction half-reaction, we need to balance the oxygen atoms by adding water molecules and balance the hydrogen atoms by adding protons (\(H^+\) ions): \(\mathrm{MnO}_{4}^{2-} + e^- \longrightarrow \mathrm{MnO}_{2} + 2H_2O + 4H^+\)

Step 6: Combine the half-reactions

Since both half-reactions involve the transfer of one electron each, we can directly add them to obtain the balanced disproportionation reaction: \(\mathrm{2MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{4}^- + \mathrm{MnO}_{2} + 2H_2O + 4H^+\)

Key Concepts

Half-Reactions in Redox

In redox chemistry, understanding half-reactions is crucial for dissecting the full story of electron exchange in a reaction. A half-reaction is either the oxidation or the reduction process in isolation, and it's a useful way to track what happens to electrons during the reaction.

A key aspect of half-reactions is they show the loss or gain of electrons explicitly. In oxidation, an atom or molecule loses electrons, represented as e-. For instance, \( \mathrm{Ni}^+ \longrightarrow \mathrm{Ni}^{2+} + e^- \) shows nickel ions (\(\mathrm{Ni}^+\)) losing an electron to form a more positively charged nickel ion (\(\mathrm{Ni}^{2+}\)). Conversely, in reduction, an atom or molecule gains electrons, such as in \( \mathrm{Ni}^+ + e^- \longrightarrow \mathrm{Ni} \) where the positively charged nickel ions (\(\mathrm{Ni}^+\)) gain an electron to form solid nickel (\(\mathrm{Ni}\)).

By breaking down complex reactions into these simpler half-reactions, we gain a clearer understanding of the electron flow and can more easily balance the overall reaction equations.

Balancing Oxidation-Reduction Equations

Balancing oxidation-reduction (redox) equations is like solving a puzzle—you need to make sure all the pieces fit perfectly. It's about ensuring that the number of atoms of each element and the charge are balanced on both sides of the equation.

To balance redox equations, we usually follow these steps:

• Write down the half-reactions for oxidation and reduction.
• Balance the number of atoms (except for oxygen and hydrogen) in each half-reaction.
• Balance the oxygen atoms by adding water molecules (\(H_2O\)) to the side lacking oxygen.
• Balance hydrogen by adding hydrogen ions (\(H^+\)) to the side lacking hydrogen atoms.
• Balance the charge by adding electrons (\(e^-\)) to the more positive side of the half-reaction.
• If necessary, multiply the half-reactions by the appropriate coefficients to ensure the same number of electrons are gained and lost between the two half-reactions.
• Add the balanced half-reactions together and cancel out common terms.

This systematic approach to balancing maintains the law of conservation of mass and charge. For example, in the disproportionation reaction for \(\mathrm{MnO}_{4}^{2-}\), we ensure that both mass and charge are in equilibrium, resulting in the balanced equation \(\mathrm{2MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{4}^- + \mathrm{MnO}_{2} + 2H_2O + 4H^+\).

Electron Transfer in Redox Reactions

Central to redox reactions is the concept of electron transfer, which is the movement of electrons from one substance to another. In these reactions, there are always two processes occurring simultaneously: oxidation is where a substance loses electrons, and reduction is where a substance gains electrons.

For a reaction to be classified as redox, this electron transfer must be present. This exchange is crucial because it links the chemical changes of the reactants to the formation of products in both half-reactions. By examining half-reactions, we can pinpoint the exact details of electron transfer. This often helps predict the outcome of the reaction and the properties of the resulting substances.

In our disproportionation reaction examples, electron transfer is what changes the oxidation states of \(\mathrm{Ni}^+\) and \(\mathrm{MnO}_{4}^{2-}\). The \(\mathrm{Ni}^+\) both loses and gains an electron, leading to two different products, \(\mathrm{Ni}^{2+}\) and solid \(\mathrm{Ni}\). Similarly, \(\mathrm{MnO}_{4}^{2-}\) is both reduced to \(\mathrm{MnO}_2\) and oxidized to \(\mathrm{MnO}_4^-\), illustrating the concept of disproportionation in action. This highlights that in some reactions, such as disproportionation, a single element can undergo oxidation and reduction simultaneously.